Question: Let $x,$ $y,$ $z$ be positive real number such that $xyz = \frac{2}{3}.$  Compute the minimum value of
\[x^2 + 6xy + 18y^2 + 12yz + 4z^2.\]
We may think of trying to apply AM-GM directly to all five terms.  Ignoring the constants, this give us a term of
\[\sqrt[5]{x^2 \cdot xy \cdot y^2 \cdot yz \cdot z^2} = \sqrt[5]{x^3 y^4 z^3}.\]This doesn't work, because the condition is $xyz = \frac{2}{3},$ so we want a power of $xyz.$  So, to get more one power of $y,$ relative to $x$ and $z,$ we split every term except $y^2$ in half:
\[\frac{x^2}{2} + \frac{x^2}{2} + 3xy + 3xy + 18y^2 + 6yz + 6yz + 2z^2 + 2z^2.\]Then by AM-GM,
\begin{align*}
&\frac{x^2}{2} + \frac{x^2}{2} + 3xy + 3xy + 18y^2 + 6yz + 6yz + 2z^2 + 2z^2 \\
&\ge 9 \sqrt[9]{\frac{x^2}{2} \cdot \frac{x^2}{2} \cdot 3xy \cdot 3xy \cdot 18y^2 \cdot 6yz \cdot 6yz \cdot 2z^2 \cdot 2z^2} \\
&= 9 \sqrt[9]{5832x^6 y^6 z^6} \\
&= 18.
\end{align*}Equality occurs when $\frac{x^2}{2} = 3xy = 18y^2 = 6yz = 2z^2.$  Along with the condition $xyz = \frac{2}{3},$ we can solve to get $x = 2,$ $y = \frac{1}{3},$ $z = 1,$ so the minimum value is $\boxed{18}.$